Related
I'm trying to select next 20 rows after top 10 rows.
select TOP 20 *
from memberform
where Row_Number over(10)
You need to use something like a CTE (Common Table Expression) and a ROW_NUMBER to define row numberings for your data set - then select from that numbered CTE for the set of rows you want:
;WITH PagingCte AS
(
SELECT
(list of columns),
RowNum = ROW_NUMBER() OVER (ORDER BY -some-column-of-yours-)
FROM
dbo.memberform
)
SELECT
(list of columns)
FROM
PagingCte
WHERE
RowNum BETWEEN 10 AND 29
In the inner ROW_NUMBER() window function, you need to define an ordering which defines how the rows are numbered (order by date, or by ID, or whatever makes sense for you).
Without an explicit ordering, there is no next 20 after the first 10 to be had..
do you mean offset clause ?
OFFSET excludes the first set of records.
OFFSET can only be used with an ORDER BY clause.
OFFSET with FETCH NEXT returns a defined window of records.
OFFSET with FETCH NEXT is great for building pagination support.
The general syntax to exclude first n records is:
SELECT column-names
FROM table-name
ORDER BY column-names
OFFSET n ROWS
Please refer to http://www.dofactory.com/sql/order-by-offset-fetch
WITH T AS
(
SELECT TOP 30 name,
row_number() OVER (ORDER BY id) AS RN
FROM memberform
ORDER BY id
)
SELECT
MAX(CASE WHEN RN <=10 THEN name END) AS Col1,
MAX(CASE WHEN RN > 10 THEN name END) AS Col2
FROM T
GROUP BY RN % 10
I have a query that returns a large (10000+ rows) dataset. I want to order by date desc, and display the first 40 results. Is there a way to run a query like this that only retrieves those 40 results without retrieving all 10000 first?
I have something like this:
select rownum, date, * from table
order by date desc
This selects all the data and orders it by date, but the rownum is not in order so it is useless for selecting only the first 40.
ROW_NUMBER() over (ORDER BY date desc) AS rowNumber
^ Will display a rownumber in order, but I can't use it in a where clause because it is a window function. I could run this:
select * from (select ROW_NUMBER() over (ORDER BY date desc) AS rowNumber,
rownum, * from table
order by date desc) where rowNumber between pageStart and pageEnd
but this is selecting all 10000 rows. How can I do this efficiently?
SELECT *
FROM (SELECT *
FROM table
ORDER BY date DESC)
WHERE rownum <= 40
will return the first 40 rows ordered by date. If there is an index on date that can be used to find these rows, and assuming statistics are up to date, Oracle should choose to use that index to identify the 40 rows that you want and then do 40 single-row lookups against the table to retrieve the rest of the data. You could throw a /*+ first_rows(40) */ hint into the inner query if you want though that shouldn't have any effect.
For a more general discussion on pagination queries and Top N queries, here's a nice discussion from Tom Kyte and a much longer AskTom discussion.
Oracle 12c has introduced a row limiting clause:
SELECT *
FROM table
ORDER BY "date" DESC
FETCH FIRST 40 ROWS ONLY;
In earlier versions you can do:
SELECT *
FROM ( SELECT *
FROM table
ORDER BY "date" DESC )
WHERE ROWNUM <= 40;
or
SELECT *
FROM ( SELECT *,
ROW_NUMBER() OVER ( ORDER BY "date" DESC ) AS RN
FROM table )
WHERE RN <= 40;
or
SELECT *
FROM TEST
WHERE ROWID IN ( SELECT ROWID
FROM ( SELECT "Date" FROM TEST ORDER BY "Date" DESC )
WHERE ROWNUM <= 40 );
Whatever you do, the database will need to look through all the values in the date column to find the 40 first items.
You don't need a window function. See
http://www.techonthenet.com/oracle/questions/top_records.php
for an answer to your problem.
I'm working on a small project in which I'll need to select a record from a temporary table based on the actual row number of the record.
How can I select a record based on its row number?
A couple of the other answers touched on the problem, but this might explain. There really isn't an order implied in SQL (set theory). So to refer to the "fifth row" requires you to introduce the concept
Select *
From
(
Select
Row_Number() Over (Order By SomeField) As RowNum
, *
From TheTable
) t2
Where RowNum = 5
In the subquery, a row number is "created" by defining the order you expect. Now the outer query is able to pull the fifth entry out of that ordered set.
Technically SQL Rows do not have "RowNumbers" in their tables. Some implementations (Oracle, I think) provide one of their own, but that's not standard and SQL Server/T-SQL does not. You can add one to the table (sort of) with an IDENTITY column.
Or you can add one (for real) in a query with the ROW_NUMBER() function, but unless you specify your own unique ORDER for the rows, the ROW_NUMBERS will be assigned non-deterministically.
What you're looking for is the row_number() function, as Kaf mentioned in the comments.
Here is an example:
WITH MyCte AS
(
SELECT employee_id,
RowNum = row_number() OVER ( order by employee_id )
FROM V_EMPLOYEE
ORDER BY Employee_ID
)
SELECT employee_id
FROM MyCte
WHERE RowNum > 0
There are 3 ways of doing this.
Suppose u have an employee table with the columns as emp_id, emp_name, salary. You need the top 10 employees who has highest salary.
Using row_number() analytic function
Select * from
( select emp_id,emp_name,row_number() over (order by salary desc) rank
from employee)
where rank<=10
Using rank() analytic function
Select * from
( select emp_id,emp_name,rank() over (order by salary desc) rank
from employee)
where rank<=10
Using rownum
select * from
(select * from employee order by salary desc)
where rownum<=10;
This will give you the rows of the table without being re-ordered by some set of values:
SELECT ROW_NUMBER() OVER (ORDER BY (SELECT '1')) AS RowID, * FROM #table
If using SQL Server 2012 you can now use offset/fetch:
declare #rowIndexToFetch int
set #rowIndexToFetch = 0
select
*
from
dbo.EntityA ea
order by
ea.Id
offset #rowIndexToFetch rows
fetch next 1 rows only
My query is as follows
BEGIN
WITH MyCTE
AS (
SELECT T.MusicAlbumTitle
,D.musicTitle
,D.mVideoID
,D.musicFileName
,T.ReleaseDate AS ReleasedDate
,D.MusicLength
,D.musicSinger
,D.MusicVideoID
,D.ExternalLink
,D.CoverImg
,ROW_NUMBER() OVER (
PARTITION BY D.MusicVideoID ORDER BY D.mVideoID
) AS row_num
FROM dbo.Music_Video T
JOIN dbo.Music_Video_Details D ON T.MusicVideoID = D.MusicVideoID
WHERE T.PortalID = #PortalID
AND T.CultureCode = #CultureCode
AND T.ComingSoon <> 1
GROUP BY T.MusicAlbumTitle
,D.musicTitle
,D.mVideoID
,T.ReleaseDate
,D.musicFileName
,D.MusicLength
,D.musicSinger
,D.MusicVideoID
,D.ExternalLink
,D.CoverImg
)
SELECT a.mVideoID
,a.MusicVideoID
,a.musicFileName
,a.MusicAlbumTitle
,a.ReleasedDate
,a.row_num
,a.CoverImg
,a.ExternalLink
,a.musicTitle
,a.MusicLength
FROM MyCTE a
WHERE row_num = 1
ORDER BY MusicVideoID DESC
END
I need to achieve total row count from last select statement.
which mean total row count that is being selected.
or any idea that might be use in this condition
How can i do this ..
Please add COUNT(*) OVER() in your select, which returns total rows selected as a new column.
Ex:
SELECT
*,
COUNT(*) OVER() AS [Total_Rows]
FROM YourTable
Just to be clear, you need to add the count to the CTE, not the outer query. The outer select is returning only one row, so the count would always be one.
The CTE should start:
WITH MyCTE
AS (
SELECT T.MusicAlbumTitle
,D.musicTitle
,D.mVideoID
,D.musicFileName
,T.ReleaseDate AS ReleasedDate
,D.MusicLength
,D.musicSinger
,D.MusicVideoID
,D.ExternalLink
,D.CoverImg
,ROW_NUMBER() OVER (
PARTITION BY D.MusicVideoID ORDER BY D.mVideoID
) AS row_num,
COUNT(*) over () as total_count
What is the simplest SQL query to find the second largest integer value in a specific column?
There are maybe duplicate values in the column.
SELECT MAX( col )
FROM table
WHERE col < ( SELECT MAX( col )
FROM table )
SELECT MAX(col)
FROM table
WHERE col NOT IN ( SELECT MAX(col)
FROM table
);
In T-Sql there are two ways:
--filter out the max
select max( col )
from [table]
where col < (
select max( col )
from [table] )
--sort top two then bottom one
select top 1 col
from (
select top 2 col
from [table]
order by col) topTwo
order by col desc
In Microsoft SQL the first way is twice as fast as the second, even if the column in question is clustered.
This is because the sort operation is relatively slow compared to the table or index scan that the max aggregation uses.
Alternatively, in Microsoft SQL 2005 and above you can use the ROW_NUMBER() function:
select col
from (
select ROW_NUMBER() over (order by col asc) as 'rowNum', col
from [table] ) withRowNum
where rowNum = 2
I see both some SQL Server specific and some MySQL specific solutions here, so you might want to clarify which database you need. Though if I had to guess I'd say SQL Server since this is trivial in MySQL.
I also see some solutions that won't work because they fail to take into account the possibility for duplicates, so be careful which ones you accept. Finally, I see a few that will work but that will make two complete scans of the table. You want to make sure the 2nd scan is only looking at 2 values.
SQL Server (pre-2012):
SELECT MIN([column]) AS [column]
FROM (
SELECT TOP 2 [column]
FROM [Table]
GROUP BY [column]
ORDER BY [column] DESC
) a
MySQL:
SELECT `column`
FROM `table`
GROUP BY `column`
ORDER BY `column` DESC
LIMIT 1,1
Update:
SQL Server 2012 now supports a much cleaner (and standard) OFFSET/FETCH syntax:
SELECT [column]
FROM [Table]
GROUP BY [column]
ORDER BY [column] DESC
OFFSET 1 ROWS
FETCH NEXT 1 ROWS ONLY;
I suppose you can do something like:
SELECT *
FROM Table
ORDER BY NumericalColumn DESC
LIMIT 1 OFFSET 1
or
SELECT *
FROM Table ORDER BY NumericalColumn DESC
LIMIT (1, 1)
depending on your database server. Hint: SQL Server doesn't do LIMIT.
The easiest would be to get the second value from this result set in the application:
SELECT DISTINCT value
FROM Table
ORDER BY value DESC
LIMIT 2
But if you must select the second value using SQL, how about:
SELECT MIN(value)
FROM ( SELECT DISTINCT value
FROM Table
ORDER BY value DESC
LIMIT 2
) AS t
you can find the second largest value of column by using the following query
SELECT *
FROM TableName a
WHERE
2 = (SELECT count(DISTINCT(b.ColumnName))
FROM TableName b WHERE
a.ColumnName <= b.ColumnName);
you can find more details on the following link
http://www.abhishekbpatel.com/2012/12/how-to-get-nth-maximum-and-minimun.html
MSSQL
SELECT *
FROM [Users]
order by UserId desc OFFSET 1 ROW
FETCH NEXT 1 ROW ONLY;
MySQL
SELECT *
FROM Users
order by UserId desc LIMIT 1 OFFSET 1
No need of sub queries ... just skip one row and select second rows after order by descending
A very simple query to find the second largest value
SELECT `Column`
FROM `Table`
ORDER BY `Column` DESC
LIMIT 1,1;
SELECT MAX(Salary)
FROM Employee
WHERE Salary NOT IN ( SELECT MAX(Salary)
FROM Employee
)
This query will return the maximum salary, from the result - which not contains maximum salary from overall table.
Old question I know, but this gave me a better exec plan:
SELECT TOP 1 LEAD(MAX (column)) OVER (ORDER BY column desc)
FROM TABLE
GROUP BY column
This is very simple code, you can try this :-
ex :
Table name = test
salary
1000
1500
1450
7500
MSSQL Code to get 2nd largest value
select salary from test order by salary desc offset 1 rows fetch next 1 rows only;
here 'offset 1 rows' means 2nd row of table and 'fetch next 1 rows only' is for show only that 1 row. if you dont use 'fetch next 1 rows only' then it shows all the rows from the second row.
Simplest of all
select sal
from salary
order by sal desc
limit 1 offset 1
select * from (select ROW_NUMBER() over (Order by Col_x desc) as Row, Col_1
from table_1)as table_new tn inner join table_1 t1
on tn.col_1 = t1.col_1
where row = 2
Hope this help to get the value for any row.....
Use this query.
SELECT MAX( colname )
FROM Tablename
where colname < (
SELECT MAX( colname )
FROM Tablename)
select min(sal) from emp where sal in
(select TOP 2 (sal) from emp order by sal desc)
Note
sal is col name
emp is table name
select col_name
from (
select dense_rank() over (order by col_name desc) as 'rank', col_name
from table_name ) withrank
where rank = 2
SELECT
*
FROM
table
WHERE
column < (SELECT max(columnq) FROM table)
ORDER BY
column DESC LIMIT 1
It is the most esiest way:
SELECT
Column name
FROM
Table name
ORDER BY
Column name DESC
LIMIT 1,1
As you mentioned duplicate values . In such case you may use DISTINCT and GROUP BY to find out second highest value
Here is a table
salary
:
GROUP BY
SELECT amount FROM salary
GROUP by amount
ORDER BY amount DESC
LIMIT 1 , 1
DISTINCT
SELECT DISTINCT amount
FROM salary
ORDER BY amount DESC
LIMIT 1 , 1
First portion of LIMIT = starting index
Second portion of LIMIT = how many value
Tom, believe this will fail when there is more than one value returned in select max([COLUMN_NAME]) from [TABLE_NAME] section. i.e. where there are more than 2 values in the data set.
Slight modification to your query will work -
select max([COLUMN_NAME])
from [TABLE_NAME]
where [COLUMN_NAME] IN ( select max([COLUMN_NAME])
from [TABLE_NAME]
)
select max(COL_NAME)
from TABLE_NAME
where COL_NAME in ( select COL_NAME
from TABLE_NAME
where COL_NAME < ( select max(COL_NAME)
from TABLE_NAME
)
);
subquery returns all values other than the largest.
select the max value from the returned list.
This is an another way to find the second largest value of a column.Consider the table 'Student' and column 'Age'.Then the query is,
select top 1 Age
from Student
where Age in ( select distinct top 2 Age
from Student order by Age desc
) order by Age asc
select age
from student
group by id having age< ( select max(age)
from student
)
order by age
limit 1
SELECT MAX(sal)
FROM emp
WHERE sal NOT IN ( SELECT top 3 sal
FROM emp order by sal desc
)
this will return the third highest sal of emp table
select max(column_name)
from table_name
where column_name not in ( select max(column_name)
from table_name
);
not in is a condition that exclude the highest value of column_name.
Reference : programmer interview
Something like this? I haven't tested it, though:
select top 1 x
from (
select top 2 distinct x
from y
order by x desc
) z
order by x
See How to select the nth row in a SQL database table?.
Sybase SQL Anywhere supports:
SELECT TOP 1 START AT 2 value from table ORDER BY value
Using a correlated query:
Select * from x x1 where 1 = (select count(*) from x where x1.a < a)
select * from emp e where 3>=(select count(distinct salary)
from emp where s.salary<=salary)
This query selects the maximum three salaries. If two emp get the same salary this does not affect the query.